Do equations of unconditional probability hold true after conditioning on both sides?Applying law of total probability to conditional probabilityHelp in Independence of eventsConditional Probability MatricesA Special Property of Finite Probability Space.Defining the set of pre-images of a product of random variables in terms of the sets of pre-image of the original random variablesMax and sum of random variablesCopula of $a_1(X_1),a_2(X_2)$Implications of inequalities involving joint probabilities on the marginalsIs the Monotone Likelihood Ratio Property (MLRP) preserved under mean-preserving spreads?Necessary and sufficient conditions on a trivariate probability distribution for being the probability distribution of $(X,Y,X-Y)$Ross, Joint distribution function with bounds on X and Y
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Do equations of unconditional probability hold true after conditioning on both sides?
Applying law of total probability to conditional probabilityHelp in Independence of eventsConditional Probability MatricesA Special Property of Finite Probability Space.Defining the set of pre-images of a product of random variables in terms of the sets of pre-image of the original random variablesMax and sum of random variablesCopula of $a_1(X_1),a_2(X_2)$Implications of inequalities involving joint probabilities on the marginalsIs the Monotone Likelihood Ratio Property (MLRP) preserved under mean-preserving spreads?Necessary and sufficient conditions on a trivariate probability distribution for being the probability distribution of $(X,Y,X-Y)$Ross, Joint distribution function with bounds on X and Y
$begingroup$
Reading a book in which the author conditions on both sides of the equation of non-conditional probability and states that they still hold after conditioning without further proving. I am wondering if this is always true.
Assume we have identity:
$$mathbbP(f(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)))$$
where $A_i, i=1,2,ldots,n$ are some events; $f,f_1,ldots,f_m$ are functions on sets using set operations including $cup$, $cap$, $setminus$;$F$ is a function defined on $[0,1]^m$. Then if I condition both sides on event X, do I have:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid X))$$
Step 2: sometimes the original equation is already conditional as below:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)))tag*$$
then do I have the (further) conditional version:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n)cap X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)cap X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)cap X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)cap X))$$
If any term of the original equation (*) is not conditional, simply use $f(...|X)$.
Step 3: Reverse the above statement, if "conditional version" holds true, how about the "unconditional version"?
If the above statements hold true, I can basically insert/remove conditional probability at will whenever I have an equation at hands, e.g.,
Since $mathbbP(Asetminus B)=mathbbP(A)-mathbbP(Acap B)$, we have $mathbbP(Asetminus Bmid X)=mathbbP(Amid X)-mathbbP(Acap Bmid X)$;
Since $mathbbP(A)=mathbbP(B_1)mathbbP(Amid B_1)+mathbbP(B_2)mathbbP(Amid B_2)$ where $B_1$ and $B_2$ is a partition of sample space, then $mathbbP(Amid X)=mathbbP(B_1mid X)mathbbP(Amid B_1 cap X)+mathbbP(B_2mid X)mathbbP(Amid B_2 cap X)$
Since $mathbbP(Amid Bcap C) = fracmathbbP(Acap Bmid C)mathbbP(Bmid C)$, we have $P(Amid B)=fracP(Acap B)P(B)$
There are many posts (like this and this) proving that the above statement is correct for a specific equation, but I haven't found a post handling the issue in a general way.
probability probability-theory conditional-probability
asked Mar 29 at 15:08
NicholasNicholas
1698
$endgroup$
add a comment |
$begingroup$
Reading a book in which the author conditions on both sides of the equation of non-conditional probability and states that they still hold after conditioning without further proving. I am wondering if this is always true.
Assume we have identity:
$$mathbbP(f(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)))$$
where $A_i, i=1,2,ldots,n$ are some events; $f,f_1,ldots,f_m$ are functions on sets using set operations including $cup$, $cap$, $setminus$;$F$ is a function defined on $[0,1]^m$. Then if I condition both sides on event X, do I have:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid X))$$
Step 2: sometimes the original equation is already conditional as below:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)))tag*$$
then do I have the (further) conditional version:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n)cap X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)cap X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)cap X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)cap X))$$
If any term of the original equation (*) is not conditional, simply use $f(...|X)$.
Step 3: Reverse the above statement, if "conditional version" holds true, how about the "unconditional version"?
If the above statements hold true, I can basically insert/remove conditional probability at will whenever I have an equation at hands, e.g.,
Since $mathbbP(Asetminus B)=mathbbP(A)-mathbbP(Acap B)$, we have $mathbbP(Asetminus Bmid X)=mathbbP(Amid X)-mathbbP(Acap Bmid X)$;
Since $mathbbP(A)=mathbbP(B_1)mathbbP(Amid B_1)+mathbbP(B_2)mathbbP(Amid B_2)$ where $B_1$ and $B_2$ is a partition of sample space, then $mathbbP(Amid X)=mathbbP(B_1mid X)mathbbP(Amid B_1 cap X)+mathbbP(B_2mid X)mathbbP(Amid B_2 cap X)$
Since $mathbbP(Amid Bcap C) = fracmathbbP(Acap Bmid C)mathbbP(Bmid C)$, we have $P(Amid B)=fracP(Acap B)P(B)$
There are many posts (like this and this) proving that the above statement is correct for a specific equation, but I haven't found a post handling the issue in a general way.
probability probability-theory conditional-probability
asked Mar 29 at 15:08
NicholasNicholas
1698
$endgroup$
$begingroup$
A conditional distribution is still a probability distribution. So any property of unconditional probabilities $mathbbP(cdot)$ holds for conditional probabilities $mathbbP(cdot mid X)$.
$endgroup$
– angryavian
Mar 29 at 17:14
$begingroup$
@angryavian So this is enough as a rigorous proof? Does it use anything like three axioms of probability, don't find it enough myself lol. Also, how about the other way around (as I mentioned in step 3)?
$endgroup$
– Nicholas
Mar 29 at 17:18
add a comment |
$begingroup$
Reading a book in which the author conditions on both sides of the equation of non-conditional probability and states that they still hold after conditioning without further proving. I am wondering if this is always true.
Assume we have identity:
$$mathbbP(f(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)))$$
where $A_i, i=1,2,ldots,n$ are some events; $f,f_1,ldots,f_m$ are functions on sets using set operations including $cup$, $cap$, $setminus$;$F$ is a function defined on $[0,1]^m$. Then if I condition both sides on event X, do I have:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid X))$$
Step 2: sometimes the original equation is already conditional as below:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)))tag*$$
then do I have the (further) conditional version:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n)cap X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)cap X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)cap X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)cap X))$$
If any term of the original equation (*) is not conditional, simply use $f(...|X)$.
Step 3: Reverse the above statement, if "conditional version" holds true, how about the "unconditional version"?
If the above statements hold true, I can basically insert/remove conditional probability at will whenever I have an equation at hands, e.g.,
Since $mathbbP(Asetminus B)=mathbbP(A)-mathbbP(Acap B)$, we have $mathbbP(Asetminus Bmid X)=mathbbP(Amid X)-mathbbP(Acap Bmid X)$;
Since $mathbbP(A)=mathbbP(B_1)mathbbP(Amid B_1)+mathbbP(B_2)mathbbP(Amid B_2)$ where $B_1$ and $B_2$ is a partition of sample space, then $mathbbP(Amid X)=mathbbP(B_1mid X)mathbbP(Amid B_1 cap X)+mathbbP(B_2mid X)mathbbP(Amid B_2 cap X)$
Since $mathbbP(Amid Bcap C) = fracmathbbP(Acap Bmid C)mathbbP(Bmid C)$, we have $P(Amid B)=fracP(Acap B)P(B)$
There are many posts (like this and this) proving that the above statement is correct for a specific equation, but I haven't found a post handling the issue in a general way.
probability probability-theory conditional-probability
asked Mar 29 at 15:08
NicholasNicholas
1698
$endgroup$
Reading a book in which the author conditions on both sides of the equation of non-conditional probability and states that they still hold after conditioning without further proving. I am wondering if this is always true.
Assume we have identity:
$$mathbbP(f(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)))$$
where $A_i, i=1,2,ldots,n$ are some events; $f,f_1,ldots,f_m$ are functions on sets using set operations including $cup$, $cap$, $setminus$;$F$ is a function defined on $[0,1]^m$. Then if I condition both sides on event X, do I have:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid X))$$
Step 2: sometimes the original equation is already conditional as below:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n))=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)))tag*$$
then do I have the (further) conditional version:
$$mathbbP(f(A_1,A_2,ldots, A_n)mid g(A_1,A_2,ldots, A_n)cap X)=F(mathbbP(f_1(A_1,A_2,ldots, A_n)mid g_1(A_1,A_2,ldots, A_n)cap X),mathbbP(f_2(A_1,A_2,ldots, A_n)mid g_2(A_1,A_2,ldots, A_n)cap X),ldots,mathbbP(f_m(A_1,A_2,ldots, A_n)mid g_m(A_1,A_2,ldots, A_n)cap X))$$
If any term of the original equation (*) is not conditional, simply use $f(...|X)$.
Step 3: Reverse the above statement, if "conditional version" holds true, how about the "unconditional version"?
If the above statements hold true, I can basically insert/remove conditional probability at will whenever I have an equation at hands, e.g.,
Since $mathbbP(Asetminus B)=mathbbP(A)-mathbbP(Acap B)$, we have $mathbbP(Asetminus Bmid X)=mathbbP(Amid X)-mathbbP(Acap Bmid X)$;
Since $mathbbP(A)=mathbbP(B_1)mathbbP(Amid B_1)+mathbbP(B_2)mathbbP(Amid B_2)$ where $B_1$ and $B_2$ is a partition of sample space, then $mathbbP(Amid X)=mathbbP(B_1mid X)mathbbP(Amid B_1 cap X)+mathbbP(B_2mid X)mathbbP(Amid B_2 cap X)$
Since $mathbbP(Amid Bcap C) = fracmathbbP(Acap Bmid C)mathbbP(Bmid C)$, we have $P(Amid B)=fracP(Acap B)P(B)$
There are many posts (like this and this) proving that the above statement is correct for a specific equation, but I haven't found a post handling the issue in a general way.
probability probability-theory conditional-probability
probability probability-theory conditional-probability
asked Mar 29 at 15:08
NicholasNicholas
1698
asked Mar 29 at 15:08
NicholasNicholas
1698
edited Mar 29 at 17:10
Nicholas
asked Mar 29 at 15:08
NicholasNicholas
1698
asked Mar 29 at 15:08
NicholasNicholas
1698
asked Mar 29 at 15:08
NicholasNicholas
1698
1698
$begingroup$
A conditional distribution is still a probability distribution. So any property of unconditional probabilities $mathbbP(cdot)$ holds for conditional probabilities $mathbbP(cdot mid X)$.
$endgroup$
– angryavian
Mar 29 at 17:14
$begingroup$
@angryavian So this is enough as a rigorous proof? Does it use anything like three axioms of probability, don't find it enough myself lol. Also, how about the other way around (as I mentioned in step 3)?
$endgroup$
– Nicholas
Mar 29 at 17:18
add a comment |
$begingroup$
A conditional distribution is still a probability distribution. So any property of unconditional probabilities $mathbbP(cdot)$ holds for conditional probabilities $mathbbP(cdot mid X)$.
$endgroup$
– angryavian
Mar 29 at 17:14
$begingroup$
@angryavian So this is enough as a rigorous proof? Does it use anything like three axioms of probability, don't find it enough myself lol. Also, how about the other way around (as I mentioned in step 3)?
$endgroup$
– Nicholas
Mar 29 at 17:18
$begingroup$
A conditional distribution is still a probability distribution. So any property of unconditional probabilities $mathbbP(cdot)$ holds for conditional probabilities $mathbbP(cdot mid X)$.
$endgroup$
– angryavian
Mar 29 at 17:14
$begingroup$
A conditional distribution is still a probability distribution. So any property of unconditional probabilities $mathbbP(cdot)$ holds for conditional probabilities $mathbbP(cdot mid X)$.
$endgroup$
– angryavian
Mar 29 at 17:14
$begingroup$
@angryavian So this is enough as a rigorous proof? Does it use anything like three axioms of probability, don't find it enough myself lol. Also, how about the other way around (as I mentioned in step 3)?
$endgroup$
– Nicholas
Mar 29 at 17:18
$begingroup$
@angryavian So this is enough as a rigorous proof? Does it use anything like three axioms of probability, don't find it enough myself lol. Also, how about the other way around (as I mentioned in step 3)?
$endgroup$
– Nicholas
Mar 29 at 17:18
add a comment |
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$begingroup$
A conditional distribution is still a probability distribution. So any property of unconditional probabilities $mathbbP(cdot)$ holds for conditional probabilities $mathbbP(cdot mid X)$.
$endgroup$
– angryavian
Mar 29 at 17:14
$begingroup$
@angryavian So this is enough as a rigorous proof? Does it use anything like three axioms of probability, don't find it enough myself lol. Also, how about the other way around (as I mentioned in step 3)?
$endgroup$
– Nicholas
Mar 29 at 17:18